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I have to prove the following inequality: $$4xy\sqrt{e^x}\sqrt{e^y}\leq (x+y)(ye^x+xe^y).$$

It's not explicitly stated that I have to use convexity as a proof however it is suggested, and I really don't understand how I'm supposed to do it I understand the definition of convexity but I'm not sure how to apply it here.

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The inequality does not necessarily hold if $x$ and $y$ are of opposite sign (consider $y=-1$ and large $x > 0$).

For $x, y> 0$ you can rewrite the inequality as $$ \frac{e^{\frac{x+y}2}}{\frac{x+y}2} \le \frac 12 \left( \frac{e^x}{x} + \frac{e^y}{y}\right) \,, $$ that hopefully helps to solve the problem using convexity.

For $x,y < 0$ you can proceed similarly.

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Consider the function $$f(x)=\frac{e^x}{x},\ x>0,$$ then $$f''(x)=\frac{(x-1)^2+1}{x^3}e^x>0, \forall x>0.$$ So $$f\left(\frac{x+y}{2}\right)\leq \frac{1}{2}(f(x)+f(y))$$ which is the desired inequality!

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Assume $xy\ge 0$ since the given inequality is not necessarily true if $x$ and $y$ have different signs. Since $2ab \le a^2+b^2$ for all $a,b\ge 0$, we find that $$ 2\sqrt{|x|}\sqrt{|y|}\le |x|+|y|, $$ $$ 2\sqrt{|y|e^x}\sqrt{|x|e^y}\le |y|e^x+|x|e^y. $$ Combining them, we obtain $$ 4|xy|\sqrt{e^x}\sqrt{e^y}\le (|x|+|y|)(|y|e^x+|x|e^y), $$ or equivalently $$ 4xy\sqrt{e^x}\sqrt{e^y}\le (x+y)(ye^x+xe^y) $$ for all $xy\ge 0$.

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